**How to Prove Points are Collinear Using Vectors :**

Here we are going to see how to prove points are collinear using vectors.

**Question 1 :**

Show that the points (2, - 1, 3), (4, 3, 1) and (3, 1, 2) are collinear.

**Solution :**

Let A (2, - 1, 3), B (4, 3, 1) and C (3, 1, 2) are collinear

= (1/2)[2(6 - 1) + 1(8 - 3) + 3(4 - 9)]

= (1/2)[ 2(5) + 1(5) + 3(-5)]

= (1/2) [10 + 5 - 15]

= (1/2)[15 - 15]

= 0

Hence the given points are collinear.

**Question 2 :**

If a vector, b vector are unit vectors and θ is the angle between them, show that

(i) sin (θ/2) = (1/2) |a vector - b vector|

**Solution :**

In the first step, let us take square for |a vector - b vector|

Since a vector and b vector are unit vectors, we have to apply 1 instead of a vector and b vector.

|a vector - b vector|^{2} = 1^{2} + 1^{2} - 2(1)(1) cos θ

|a vector - b vector|^{2} = 2 - 2 cosθ

|a vector - b vector|^{2} = 2(1- cosθ)

|a vector - b vector|^{2} = 2 (2 sin^{2 }(θ/2))

|a vector - b vector| = 2 sin (θ/2)

sin (θ/2) = (1/2)|a vector - b vector| --(1)

Hence it is proved

(ii) cos (θ/2) = (1/2) |a vector + b vector|

**Solution :**

In the first step, let us take square for |a vector + b vector|

Since a vector and b vector are unit vectors, we have to apply 1 instead of a vector and b vector.

|a vector + b vector|^{2} = 1^{2} + 1^{2} + 2(1)(1) cos θ

|a vector - b vector|^{2} = 2 + 2 cosθ

|a vector - b vector|^{2} = 2(1 + cosθ)

|a vector - b vector|^{2} = 2 (2 cos^{2 }(θ/2))

|a vector - b vector| = 2 cos (θ/2)

cos (θ/2) = (1/2)|a vector + b vector| --(2)

Hence it is proved

(ii) tan (θ/2) = |a vector-b vector| / |a vector-b vector|

**Solution :**

We may prove the above result using

sin (θ/2) = (1/2)|a vector - b vector| --(1)

cos (θ/2) = (1/2)|a vector + b vector| --(2)

(1)/(2) ==>

sin (θ/2)/cos (θ/2)

= (1/2)|a vector - b vector| / (1/2)|a vector + b vector|

tan (θ/2)

= (1/2)|a vector-b vector|/(1/2)|a vector+b vector|

**Question 3 :**

Let a vector, b vector, c vector be three vectors such that |a vector| = 3, |b vector|= 4, |c vector|= 5 and each one of them being perpendicular to the sum of the other two, find | a vector + b vector + c vector |

**Solution :**

| a vector + b vector + c vector | =

|a vector|^{2}+|b vector|^{2}+|c vector|^{2}+2|a vector||b vector| + 2|b vector||c vector| + 2|c vector||a vector|

= |avector|^{2}+|bvector|^{2}+|cvector|^{2}+2(a.b)+2(b.c)+ 2(c.a)

Since they are perpendicular to each other, the value of a. b, b.c and c.a will be equal to 0.

= 3^{2} + 4^{2} + 5^{2} + 2(0) + 2(0) + 2(0)

= 9 + 16 + 25

= 50

After having gone through the stuff given above, we hope that the students would have understood,"How to Prove Points are Collinear Using Vectors"

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